JEE Advanced (Single Correct Type): Mathematical induction & Binomial Theorem | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. If xⁿ – 1 is divisible by x – k, then the least positive integral value of k is
(a) 1
(b) 2
(c) 3
(d) 4

Correct Answer is option (a)
Given,
P(n): xn – 1 is divisible by x – k
Let us substitute n = 1, 2, 3,..
⇒ P(1) : x – 1
⇒ P(2) : x² – 1 = (x−1)(x+1)
⇒ P(3) : x³ – 1 = (x − 1)(x² + x + 1)
⇒ P(4) : x⁴ − 1 = (x² − 1)(x² + 1) = (x − 1)(x + 1)(x² + 1)
Therefore, the least positive integral value of k is 1.

Q.2. The coefficient of the middle term in the expansion of (2+3x)⁴ is:
(a) 5!
(b) 6
(c) 216
(d) 8!

Correct Answer is option (c)
If the exponent of the expression is n, then the total number of terms is n+1.
Hence, the total number of terms is 4+1 = 5.
Hence, the middle term is the 3rd term.
We know that general term of (x+a)ⁿ is T_r+1 =ⁿC_r x^n-r a^r
Here, n=4, r=2
Therefore, T₃ = ⁴C₂.(2)².(3x)²
T₃ = (6).(4).(9x²)
T₃ = 216x².
Therefore, the coefficient of the middle term is 216.

Q.3. For any natural number n, 2²ⁿ – 1 is divisible b
(a) 2
(b) 3
(c) 4
(d) 5

Correct Answer is option (b)
Let P(n) = 2²ⁿ – 1
Substituting n = 1, 2, 3,….
P(1) = 2²⁽¹⁾ – 1 = 4 – 1 = 3
This is divisible by 3.
P(2) = 2²⁽²⁾ – 1 = 16 – 1 = 15
This is divisible by 3.
P(3) = 2²⁽³⁾ – 1 = 256 – 1 = 255
This is also divisible by 3.
Assume that P(n) is true for some natural number k, i.e., P(k): 2^2k – 1 is divisible by 3, i.e., 2^2k – 1
= 3q, where q ∈ N
Now,
P(k + 1) : 2^2(k+1) – 1
= 2^{2k + 2} – 1
= 2^2k . 2² – 1
= 2^2k . 4 – 1
= 3.2^2k + (2^2k – 1)
= 3.2^2k + 3q
= 3 (2^2k + q) = 3m, where m ∈ N
Thus P(k + 1) is true, whenever P(k) is true.
Therefore, for any natural number n, 2²ⁿ – 1 is divisible by 3.

Q.4. The value of (126)^1/3 up to three decimal places is
(a) 5.011
(b) 5.012
(c) 5.013
(d) 5.014

Correct Answer is option (c)
(126)^⅓ can also be written as the cube root of 126.
Hence, (126)^⅓ is approximately equal to 5.013.
Hence, option (c) 5.013 is the correct answer.

Q.5. A student was asked to prove a statement P(n) by induction. He proved that P(k + 1) is true whenever P(k) is true for all k > 5 ∈ N and also that P(5) is true. Based on this, he could conclude that P(n) is true
(a) for all n ∈ N
(b) for all n > 5
(c) for all n ≥ 5
(d) for all n < 5

Correct Answer is option (c)
The student could be able to conclude that P(n) is true for all n ≥ 5 since P(5) is true for all k > 5 ∈ N as well as true for P(5) and P(k + 1) is true, whenever P(k) is true.

Q.6. If n is even in the expansion of (a+b)ⁿ, the middle term is:
(a) n^th term
(b) (n/2)^th term
(c) [(n/2)-1]^th term
(d) [(n/2)+1]^th term

Correct Answer is option (d)
In general, if “n” is the even in the expansion of (a+b)ⁿ, then the number of terms will be odd. (i.e) n+1.
Hence, the middle term of the expansion (a+b)ⁿ is [(n/2)+1]^th term.

Q.7. If P (n): “49ⁿ + 16ⁿ + k is divisible by 64 for n ∈ N” is true, then the least negative integral value of k is
(a) 1
(b) -2
(c) -1
(d) -3

Correct Answer is option (c)
Given that P(n) : 49ⁿ + 16ⁿ + k is divisible by 64 for n ∈ N
For n = 1,
P(1) : 49 + 16 + k = 65 + k is divisible by 64.
Thus k, should be -1 since, 65 – 1 = 64 is divisible by 64.

Q.8. The largest coefficient in the expansion of (1 + x)¹⁰ is:
(a) 10! / (5!)²
(b) 10! / 5!
(c) 10! / (5!×4!)²
(d) 10! / (5!×4!)

Correct Answer is option (a)
Given: (1+x)¹⁰
The greatest coefficient will always occur in the middle term.
Hence, the total number of terms in an expansion is 11. (i.e. 10+1 = 11)
Therefore, middle term = [(10/2) + 1] = 5+1 = 6th term.
We know that general term of (x+a)ⁿ is T_r+1 =ⁿC_r x^n-r a^r
Here, n=10, r=5
So, T6 = ¹⁰C₅.x⁵
Therefore, the coefficient of the greatest term = ¹⁰C₅ = 10!/(5!)².
So, option (a) 10!/(5!)² is the correct answer.

Q.9. For all n ∈ N, 3.5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by
(a) 19
(b) 17
(c) 23
(d) 25

Correct Answer is option (b)
Let P(n) be the statement that 3.5^{2n + 1} + 2^{3n + 1} is divisible by 17
If n = 1, then given expression = 3 * 5³ + 2⁴ + 375 + 16 = 391 = 17 * 23, divisible by 17.
P(1) is true
Assume that P(k) is true.
3.5^{2k + 1} + 2^{3k + 1} is divisible by 17.
3.5^2k = 1 + 2^{3k + 1} = 17m where m ∈ N
3.5^{2(k + 1) + 1} + 23^{(k + 1) + 1}
= 3.5^{2k + 1} * 5² + 2^{3k + 1} * 2³
= 25^{(17m – 23k+1)} + 8.2^{3k + 1}
= 425m – 25.2^{3k + 1} + 8.2^{3k + 1}
= 425m – 17.2^{3k + 1}
= 17(25m – 2^{3k + 1}), divisible by 17
P(k + 1) is true by Principle of Mathematical Induction
P(n) is true for all n ∈ N. 3.5^{2n + 1} + 2^{3n + 1} is divisible by 17 for all n ∈ N

Q.10. The coefficient of x³y⁴ in (2x+3y²)⁵ is
(a) 360
(b) 720
(c) 240
(d) 1080

Correct Answer is option (b)
Given: (2x+3y²)⁵
Therefore, the general form for the expression (2x+3y²)⁵ is T_r+1 = ⁵C_r. (2x)^r.(3y²)^5-r
Hence, T₃₊₁ = ⁵C₃ (2x)³.(3y²)^5-3
T₄ = ⁵C₃ (2x)³.(3y²)²
T₄ = ⁵C₃.8x³.9y⁴
On simplification, we get
T₄ = 720x³y⁴
Therefore, the coefficient of x³y⁴ in (2x+3y²)⁵ is 720.

Q.11. n(n + 1) (n + 5) is a multiple of
(a) 3
(b) 8
(c) 5
(d) 7

Correct Answer is option (a)
Let P(n) = n(n + 1)(n + 5)
Substituting n = 1, 2, 3,….
P(1) = 1(1 + 1)(1 + 5) = 2(6) = 12; multiple of 2, 3, 4, 6
P(2) = 2(2 + 1)(2 + 5) = 2(3)(7) = 42; multiple of 2, 3, 6, 7
P(3) = 3(3 + 1)(3 + 5) = 3(4)(8) = 96; multiple of 2, 3, 4, 6, 8, 12
..
Thus, from the above statements and verifying the options, we can say that n(n + 1)(n + 5) is a multiple of 3.

Q.12. The largest term in the expansion of (3+2x)⁵⁰, when x = ⅕ is
(a) 6^th term
(b) 7^th term

(c) 8^th term
(d) None of the above

Correct Answer is option (a)
The greatest term in the expansion of (x + y)ⁿ is the kth term. Where k = [(n + 1)y]/[x + y]..(1)
On comparing the given expression with the general form, x = 3, y=2x, n=50
Now, substitute the values in the given expression, we get
Hence, kth term = [(50 + 1)(2x)]/[3 + 2x]
When x =  ⅕,
K^th term = [(51)(2(⅕))]/[3 + 2(⅕)] = 6
Hence, the 6th term is the largest term in the expansion of (3 + 2x)⁵⁰, when x = ⅕.

Q.13. n² < 2ⁿ for all natural numbers
(a) n ≥ 5
(b) n < 5
(c) n > 1
(d) n ≤ 3

Correct Answer is option (a)
Consider, P(n) : n² < 2ⁿ
Substituting n = 1, 2, 3,…
P(1): 1² < 2¹
1 < 2 (not true)
P(2): 2² < 2²
4 < 4 (not true)
P(3): 3² < 2³
9 < 8 (not true)
P(4): 4² < 2⁴
16 < 16 (not true)
P(5): 5² < 2⁵
25 < 32 (true)
P(6): 6² < 2⁶
26 < 64 (true)
Thus, n² < 2ⁿ for all natural numbers n ≥ 5.

Q.14. The coefficient of y in the expansion of (y²+(c/y))⁵ is:
(a) 10c
(b) 29c
(c) 10c³
(d) 20c³

Correct Answer is option (c)
Given: (y²+(c/y))⁵
We know that general term of (x+a)ⁿ is T_r+1 =ⁿC_r x^n-r a^r
Here, n=5, r=?
(y²+(c/y))⁵ = ⁵C_r.(y²)^r.(c/y)^5-r
(y²+(c/y))⁵ = 5Cr. y^2r. (c^5-r/y^5-r)
On solving this, we get r = 2.
Hence, the coefficient of y = ⁵C₃.c³ = 10c³.
Therefore, option (c) 10c³ is the correct answer.

Q.15. If 10ⁿ + 3.4ⁿ⁺² + k is divisible by 9 for all n ∈ N, then the least positive integral value of k is
(a) 5
(b) 3
(c) 7
(d) 1

Correct Answer is option (a)
Given that 10ⁿ + 3.4ⁿ⁺² + k is exactly divisible by 9.
Consider: P(n) = 10ⁿ + 3.4ⁿ⁺² + k
Substituting n = 1,
P(1) = 10¹ + 3.4¹⁺² + k
= 10 + 3(64) + k
= 10 + 192 + k
= 202 + k is exactly divisible by 9, the value of k will be 5.

Q.16. The fourth term in the expansion of (x-2y)¹² is:
(a) -1760 x⁹ × y³
(b) -1670 x⁹ × y³
(c) -7160 x⁹ × y³
(d) -1607 x⁹ × y³

Correct Answer is option (a)
We know that the general term of an expansion (a+b)ⁿ is T_r+1 = ⁿC_r a^n-r b^r.
Now, we have to find the fourth term in the expansion (x-2y)¹²
Hence, r = 3, a = x, b = -2y, n= 12.
Now, substitute the values in the formula, we get
T₃₊₁ = ¹²C₃ x^12-3 (-2y)³.
On solving this, we get
T₄ = -1760x⁹y³.

Q.17. Let P(n) : “2ⁿ < (1 × 2 × 3 × … × n)”. Then the smallest positive integer for which P(n) is true is
(a) 1
(b) 2
(c) 3
(d) 4

Correct Answer is option (d)
P(1) : 2¹ < 1
2< 1 is false
P(2) : 2² < 1 × 2
4 < 2 is false
P(3) : 2³ < 1 × 2 × 3
8 < 6 is false
P(4) : 2⁴ < 1 × 2 × 3 × 4
16 < 24 is true

Q.18. If the fourth term of the binomial expansion of (px+(1/x))ⁿ is 5/2, then
(a) n=6, p=6
(b) n=8, p=6
(c) n=8, p= ½
(d) n=6, p=½

Correct Answer is option (d)
Given: (px+(1/x))ⁿ
Hence, the fourth term, T₃₊₁ = ⁿC₃(px)^n-3(1/x)³
Given that the fourth term of the binomial expansion of (px+(1/x))ⁿ is 5/2, which is independent of x.
Hence, (5/2)= ⁿC₃(px)^n-3(1/x)³ …(1)
On solving this, we get n=6.
Now, substitute n=6 in (5/2)= ⁿC₃(p)³
20p³ = 5/2
p³=⅛
p=½.
Therefore, n=6 and p=½.

Q.19. For every positive integer n, 7n – 3n is divisible by
(a) 3
(b) 4
(c) 7
(d) 5

Correct Answer is option (b)
Let P(n) = 7ⁿ – 3ⁿ
Substituting n = 1, 2, 3,…
P(1) = 7¹ – 3¹ = 7 – 3 = 4
P(2) = 7² – 3² = 49 – 9 = 40
P(3) = 7³ – 3³ = 343 – 27 = 316
Thus, for every positive integer n, 7ⁿ – 3ⁿ is divisible by 4.

Q.20. If n is the positive integer, then 2³ⁿ – 7n -1 is divisible by
(a) 7
(b) 10
(c) 49
(d) 81

Correct Answer is option (c)
Given: 2³ⁿ – 7n -1. It can also be written as 8ⁿ – 7n – 1
Let 8ⁿ – 7n – 1 =0
So, 8ⁿ = 7n+1
8ⁿ = (1+7)ⁿ
By applying binomial theorem, we get
8n – 1 – 7n = 49 (or) 2³ⁿ – 7n -1 = 49
Hence, 2³ⁿ – 7n -1 is divisible by 49.